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2 years ago

Find the slope of the line that passes through (5, 5) and (1, 0),

Mathematics

1 answer:

Ludmilka [50]2 years ago

3 0

Answer:

Slope is 5/4.

Step-by-step explanation:

The slope of two points can be expressed as $\displaystyle \large{m=\frac{y_2-y_1}{x_2-x_1}$ where m represents the slope,

Let (5,5) be (x1,y1) and (1,0) be (x2,y2). Thus:

$\displaystyle \large{m=\frac{0-5}{1-5}=\frac{-5}{-4}=\frac{5}{4}}$

You can also consider both points other way around. We can let (1,0) be (x1,y1) and (5,5) be (x2,y2) and therefore:

$\displaystyle \large{m=\frac{5-0}{5-1}=\frac{5}{4}}$

Therefore, the slope is 5/4.

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Can someone help me with this

jolli1 [7]

Answer:

4 sides

Step-by-step explanation:

A lot of math is about pattern matching. Here, you're asked if sides are the same length (one side matches another), if there are right angles (square corners), and if the number of angles or sides is 4.

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pink: 2 pairs of parallel sides that are the same length (4 sides). Adjacent sides are different length.

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2 years ago

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antiseptic1488 [7]

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Rate = 2.5%

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2 years ago

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Guys! Help me find how much of this circle is shaded!

Alona [7]

Answer:

The answer is 1/10

Step-by-step explanation:

Let us take the Shaded region be x

So,

2/5 + x = 1/2

Now, Solve for x

x + 2/5 = 1/2

5x + 2/5 = 1/2

2(5x + 2) = 5(1)

10x + 4 = 5

10x + 4 – 4 = 5 – 4

10x= 1

10x/10 = 1/10

x = 1/10

Thus, The value of x is 1/10

<h3>For Verification;</h3>

2/5 + x = 1/2

2/5 + 1/10 = 1/2

2 × 2/5 × 2 + 1/10 = 1/2

4/10 + 1/10 = 1/2

4 + 1/10 = 1/2

5/10 = 1/2

1/2 = 1/2

L.H.S = R.H.S

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-TheUnknownScientist 72

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2 years ago

Solve the equation <img src="https://tex.z-dn.net/?f=%2835%20x%5E%7B4%7D%20y%2B14%20x%5E%7B5%7D%20y-2%20y%5E%7B3%7D-4x%20y%5E%7B

jeka94

$\underbrace{35x^4y+14x^5y-2y^3-4xy^3}_M\,\mathrm dx+\underbrace{7x^5-6xy^2}_N\,\mathrm dy=0$

$M_y=35x^4+14x^5-6y^2-12xy^2$
$N_x=35x^4-6y^2$

$\dfrac{N_x-M_y}N=\dfrac{-14x^5+12xy^2}{7x^5-6xy^2}=-2$

This suggests an integrating factor depending on $x$ only is possible, and given by

$\mu(x)=\exp\left(-\displaystyle\int\frac{N_x-M_y}N\,\mathrm dx\right)=e^{2x}$

Distributing across the ODE, we end up with

$\underbrace{(35x^4y+14x^5y-2y^3-4xy^3)e^{2x}}_{M^*}\,\mathrm dx+\underbrace{(7x^5-6xy^2)e^{2x}}_{N^*}\,\mathrm dy=0$

The equation is now exact, with

${M^*}_y={N^*}_x=(35x^4+14x^5-6y^2-12xy^2)e^{2x}$

Now we find the solution:

$F_x=M^*$
$F=\displaystyle\int(35x^4+14x^5-2y^3-4xy^3)e^{2x}\,\mathrm dx$
$F=(7x^5y-2xy^3)e^{2x}+f(y)$

$F_y=N^*$
$(7x^5-6xy^2)e^{2x}+f'(y)=(7x^5-6xy^2)e^{2x}$
$f'(y)=0$
$\implies f(y)=C$

The general solution is then

$F(x,y)=(7x^5y-2xy^3)e^{2x}=C$

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